Given the action (note $G_{ab}$ is a symmetric matrix, i.e. $G_{ba} = G_{ab}$):

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)\Big)$$

Show (using Euler Lagrange's equation) that the following equation holds:

$$\ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c = -\sum_{a} F^{da}\partial_a V$$

Where $F^{ab}$ is the inverse of $G_{ab}$.

Also note that:

$$\sum_{b} F^{ab}G_{ab} = \delta_c^a$$

$$\partial_{a} = \frac{\partial}{\partial q_a}$$

What I have done:

We know that the action functional corresponds to the Lagrangian (for the time interval $[t_0, t_1]$):

$$S[q] = \int_{t_0}^{t_1} L(q, \dot q, t)dt$$


$$L = \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)$$

Euler Lagrange's equation is:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{\partial L}{\partial q_k}$$

Let's go step by step:

1) We compute the term $\frac{\partial L}{\partial \dot q_k}$ (which turns out to be the definition of generalized momentum):

$$p_k = \frac{\partial L}{\partial \dot q_k} = \sum_{ab} \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = \sum_{b} G_{kb} \dot q^b + \sum_{a} G_{ak} \dot q^a = 2\sum_a G_{ak} \dot q^a$$

2) We now compute the term $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big)$

NOTE: I know that the symmetric matrix $G_{ab}$ only depends on $q_k$. By the chain rule (for the sake of clarity: $G_{ab} (q_k)$ notation means that the matrix $G_{ab}$ is a function of $q_k$):

$$\frac{d}{dt} \sum_a G_{ab} (q_k) = \sum_{a} \partial_k G_{ab} \dot q_k$$

That being said, let's go through the calculation:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{d}{dt}\Big( 2\sum_a G_{ak} (q_k) \dot q^a \Big) = 2\sum_a G_{ak} \ddot q^a + 2\sum_a \partial_k G_{ak} \dot q^a \dot q^k$$

3) We now compute the term $\frac{\partial L}{\partial q_k}$

$$\frac{\partial L}{\partial q_k} = \sum_{ab} \partial_k G_{ab} \dot q^a\dot q^b - \partial_k V(q)$$

Here's where I get stuck: I do not see why the following holds:

$$2\sum_a F^{da}G_{ak} \ddot q^a + 2\sum_a F^{da}\partial_k G_{ak} \dot q^a \dot q^k -\sum_{ab} F^{da}\partial_k G_{ab} \dot q^a\dot q^b + F^{da}\partial_k V(q) = \ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c + \sum_{a} F^{da}\partial_a V$$

Please let me know if need more details in the above equation; I can provide more steps.

  • $\begingroup$ Does $G_{ab}$ have an explicit expression? $\endgroup$ Dec 9, 2019 at 16:29
  • 2
    $\begingroup$ This might be helfpul: physics.stackexchange.com/q/137422 . To simplify things, also check out en.wikipedia.org/wiki/Christoffel_symbols, specifically those of the second kind. $\endgroup$
    – R. Romero
    Dec 9, 2019 at 16:45
  • 1
    $\begingroup$ Modify Newton's First law of motion to be : Objects travel along geodesics in space-time in the absence of a net force, then derive the geodesic equations. The techniques used there can be used to re derive laws of motion in the presence of a force my modifying the Lagrangian appropriately: people.uncw.edu/hermanr/GR/geodesic.pdf $\endgroup$
    – R. Romero
    Dec 9, 2019 at 16:49
  • $\begingroup$ @Tesseract all I know about $G_{ab}$ is that it is symmetric and depends on $q_{k}$ $\endgroup$
    – JD_PM
    Dec 9, 2019 at 17:05

1 Answer 1


A hint: when you use the chain rule, you need a new index, so $\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial q^k}\right)$ would be $$2\sum_aG_{ak}\ddot{q}^a+2\sum_{ac}\partial_cG_{ak}\dot{q}^a\dot{q}^c$$ We can split this up into parts, and relabel indices to get $$2\sum_aG_{ak}\ddot{q}^a+\sum_{ac}\partial_cG_{ak}\dot{q}^a\dot{q}^c+\sum_{ac}\partial_aG_{ck}\dot{q}^a\dot{q}^c$$ Can you continue from here? Also, check out this article on Einstein summation notation; it will make your work cleaner and more readable!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.